Questions on Showing $X_{n}Y_{n}xrightarrow{P} XY$
$begingroup$
Let $Y_{n}xrightarrow{P}Y$ and $X_{n}xrightarrow{P}X$ on a prob. space.$(Omega,mathcal{F},P)$. As a hint, use $X_{n}Y_{n}-XY=(X_{n}-X)(Y_{n}-Y)+X(Y_{n}-Y)+Y(X_{n}-X)$.
My question pertains only to reasoning for $X(Y_{n}-Y)xrightarrow{P}0$
Let $epsilon > 0$ and $n > 0$
$P(|X(Y_{n}-Y)|>epsilon)=P(|X||Y_{n}-Y|>epsilon)=P(|X||Y_{n}-Y|>epsilon,|X|leq n)+P(|X||Y_{n}-Y|>epsilon,|X|> n)(*)$
Everything is fine until this point, however, I do not understand the next inequality, namely:
$(*)leq P(n|Y_{n}-Y|>epsilon,|X|=n)+P(|X|>n)(**)$
Then we go on to say,
$(**)leq P(n|Y_{n}-Y|>epsilon)+P(|X|>n)=P(|Y_{n}-Y|>frac{epsilon}{n})+P(|X|>n)(***)$ and since $Y_{n} xrightarrow{P} Y$
$(***)to P(|X|>n)$
The solution then immediately implies:
$limsup_{nto infty}P(|X||Y_{n}-Y|>epsilon)=0$ which I do not understand in two parts:
$1.$ Why are we using $limsup_{nto infty}$ all of a sudden rather than $lim_{nto infty}$? I have only ever used $lim_{nto infty}$ on probability distributions.
$2.$ Our estimations must imply that $limsup_{nto infty} P(|X|>n)=0$ but what is the exact reasoning behind this. I mean intuitively it makes sense, given that ${|X|>n}$ is increasingly likely to have measure $0$ as $n to infty$ if $X$ has certain characteristics, but we know nothing about $X$ other than the fact that $X_{n} xrightarrow{P} X$.
If anyone could answer $(*),1.$ and $2.$ I would be extremely grateful.
probability probability-theory probability-distributions convergence random-variables
asked Jan 9 at 22:34
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
Let $Y_{n}xrightarrow{P}Y$ and $X_{n}xrightarrow{P}X$ on a prob. space.$(Omega,mathcal{F},P)$. As a hint, use $X_{n}Y_{n}-XY=(X_{n}-X)(Y_{n}-Y)+X(Y_{n}-Y)+Y(X_{n}-X)$.
My question pertains only to reasoning for $X(Y_{n}-Y)xrightarrow{P}0$
Let $epsilon > 0$ and $n > 0$
$P(|X(Y_{n}-Y)|>epsilon)=P(|X||Y_{n}-Y|>epsilon)=P(|X||Y_{n}-Y|>epsilon,|X|leq n)+P(|X||Y_{n}-Y|>epsilon,|X|> n)(*)$
Everything is fine until this point, however, I do not understand the next inequality, namely:
$(*)leq P(n|Y_{n}-Y|>epsilon,|X|=n)+P(|X|>n)(**)$
Then we go on to say,
$(**)leq P(n|Y_{n}-Y|>epsilon)+P(|X|>n)=P(|Y_{n}-Y|>frac{epsilon}{n})+P(|X|>n)(***)$ and since $Y_{n} xrightarrow{P} Y$
$(***)to P(|X|>n)$
The solution then immediately implies:
$limsup_{nto infty}P(|X||Y_{n}-Y|>epsilon)=0$ which I do not understand in two parts:
$1.$ Why are we using $limsup_{nto infty}$ all of a sudden rather than $lim_{nto infty}$? I have only ever used $lim_{nto infty}$ on probability distributions.
$2.$ Our estimations must imply that $limsup_{nto infty} P(|X|>n)=0$ but what is the exact reasoning behind this. I mean intuitively it makes sense, given that ${|X|>n}$ is increasingly likely to have measure $0$ as $n to infty$ if $X$ has certain characteristics, but we know nothing about $X$ other than the fact that $X_{n} xrightarrow{P} X$.
If anyone could answer $(*),1.$ and $2.$ I would be extremely grateful.
probability probability-theory probability-distributions convergence random-variables
asked Jan 9 at 22:34
SABOYSABOY
656311
$endgroup$
add a comment |
$begingroup$
Let $Y_{n}xrightarrow{P}Y$ and $X_{n}xrightarrow{P}X$ on a prob. space.$(Omega,mathcal{F},P)$. As a hint, use $X_{n}Y_{n}-XY=(X_{n}-X)(Y_{n}-Y)+X(Y_{n}-Y)+Y(X_{n}-X)$.
My question pertains only to reasoning for $X(Y_{n}-Y)xrightarrow{P}0$
Let $epsilon > 0$ and $n > 0$
$P(|X(Y_{n}-Y)|>epsilon)=P(|X||Y_{n}-Y|>epsilon)=P(|X||Y_{n}-Y|>epsilon,|X|leq n)+P(|X||Y_{n}-Y|>epsilon,|X|> n)(*)$
Everything is fine until this point, however, I do not understand the next inequality, namely:
$(*)leq P(n|Y_{n}-Y|>epsilon,|X|=n)+P(|X|>n)(**)$
Then we go on to say,
$(**)leq P(n|Y_{n}-Y|>epsilon)+P(|X|>n)=P(|Y_{n}-Y|>frac{epsilon}{n})+P(|X|>n)(***)$ and since $Y_{n} xrightarrow{P} Y$
$(***)to P(|X|>n)$
The solution then immediately implies:
$limsup_{nto infty}P(|X||Y_{n}-Y|>epsilon)=0$ which I do not understand in two parts:
$1.$ Why are we using $limsup_{nto infty}$ all of a sudden rather than $lim_{nto infty}$? I have only ever used $lim_{nto infty}$ on probability distributions.
$2.$ Our estimations must imply that $limsup_{nto infty} P(|X|>n)=0$ but what is the exact reasoning behind this. I mean intuitively it makes sense, given that ${|X|>n}$ is increasingly likely to have measure $0$ as $n to infty$ if $X$ has certain characteristics, but we know nothing about $X$ other than the fact that $X_{n} xrightarrow{P} X$.
If anyone could answer $(*),1.$ and $2.$ I would be extremely grateful.
probability probability-theory probability-distributions convergence random-variables
asked Jan 9 at 22:34
SABOYSABOY
656311
$endgroup$
Let $Y_{n}xrightarrow{P}Y$ and $X_{n}xrightarrow{P}X$ on a prob. space.$(Omega,mathcal{F},P)$. As a hint, use $X_{n}Y_{n}-XY=(X_{n}-X)(Y_{n}-Y)+X(Y_{n}-Y)+Y(X_{n}-X)$.
My question pertains only to reasoning for $X(Y_{n}-Y)xrightarrow{P}0$
Let $epsilon > 0$ and $n > 0$
$P(|X(Y_{n}-Y)|>epsilon)=P(|X||Y_{n}-Y|>epsilon)=P(|X||Y_{n}-Y|>epsilon,|X|leq n)+P(|X||Y_{n}-Y|>epsilon,|X|> n)(*)$
Everything is fine until this point, however, I do not understand the next inequality, namely:
$(*)leq P(n|Y_{n}-Y|>epsilon,|X|=n)+P(|X|>n)(**)$
Then we go on to say,
$(**)leq P(n|Y_{n}-Y|>epsilon)+P(|X|>n)=P(|Y_{n}-Y|>frac{epsilon}{n})+P(|X|>n)(***)$ and since $Y_{n} xrightarrow{P} Y$
$(***)to P(|X|>n)$
The solution then immediately implies:
$limsup_{nto infty}P(|X||Y_{n}-Y|>epsilon)=0$ which I do not understand in two parts:
$1.$ Why are we using $limsup_{nto infty}$ all of a sudden rather than $lim_{nto infty}$? I have only ever used $lim_{nto infty}$ on probability distributions.
$2.$ Our estimations must imply that $limsup_{nto infty} P(|X|>n)=0$ but what is the exact reasoning behind this. I mean intuitively it makes sense, given that ${|X|>n}$ is increasingly likely to have measure $0$ as $n to infty$ if $X$ has certain characteristics, but we know nothing about $X$ other than the fact that $X_{n} xrightarrow{P} X$.
If anyone could answer $(*),1.$ and $2.$ I would be extremely grateful.
probability probability-theory probability-distributions convergence random-variables
probability probability-theory probability-distributions convergence random-variables
asked Jan 9 at 22:34
SABOYSABOY
656311
asked Jan 9 at 22:34
SABOYSABOY
656311
asked Jan 9 at 22:34
SABOYSABOY
656311
asked Jan 9 at 22:34
SABOYSABOY
656311
asked Jan 9 at 22:34
SABOYSABOY
656311
656311
add a comment |
add a comment |
1 Answer
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$begingroup$
We use a $limsup$ because we do not know whether the limit exists, whereas $limsup$ always exists. Note that since your sequence is nonnegative, it goes to zero iff zero is its limsup.
$X$ is a random variable that is as finite. Thus $P(|X| > n)$ is a decreasing sequence with limit $P( forall,n |X| > n)=P(|X|=infty)=0$.
$(*) leq (**)$ is false. However, $(*) leq (***)$ holds.
This is because the event $(|X||Y_n-Y| > epsilon, |X| leq p)$ is contained in the event $(p|Y_n-Y| > epsilon)$, and because the event $(|X||Y_n-Y| > epsilon, |X| > p$) is contained in the event $(|X| > p)$.
Note that for the proof to work, your lower bound on $X$ should be different from the index in the sequence $Y$. The argument reads as follows:
By the same computations, $P(|X||Y_n-Y| > epsilon) leq P(|X| > p)+P(|Y_n-Y| > epsilon/p)$ for every $n,p$ thus the limsup of the LHS is not greater than $P(|X| >p)$ for every $p$. Since this can be arbitrarily close to $0$, the limsup, hence the limit, is $0$.
answered Jan 9 at 22:55
MindlackMindlack
4,770210
$endgroup$
$begingroup$
So to prove $xrightarrow{P}$ should I always use $limsup_{nto infty}$ rather than $lim_{nto infty}$
$endgroup$
– SABOY
Jan 10 at 11:59
add a comment |
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$begingroup$
We use a $limsup$ because we do not know whether the limit exists, whereas $limsup$ always exists. Note that since your sequence is nonnegative, it goes to zero iff zero is its limsup.
$X$ is a random variable that is as finite. Thus $P(|X| > n)$ is a decreasing sequence with limit $P( forall,n |X| > n)=P(|X|=infty)=0$.
$(*) leq (**)$ is false. However, $(*) leq (***)$ holds.
This is because the event $(|X||Y_n-Y| > epsilon, |X| leq p)$ is contained in the event $(p|Y_n-Y| > epsilon)$, and because the event $(|X||Y_n-Y| > epsilon, |X| > p$) is contained in the event $(|X| > p)$.
Note that for the proof to work, your lower bound on $X$ should be different from the index in the sequence $Y$. The argument reads as follows:
By the same computations, $P(|X||Y_n-Y| > epsilon) leq P(|X| > p)+P(|Y_n-Y| > epsilon/p)$ for every $n,p$ thus the limsup of the LHS is not greater than $P(|X| >p)$ for every $p$. Since this can be arbitrarily close to $0$, the limsup, hence the limit, is $0$.
answered Jan 9 at 22:55
MindlackMindlack
4,770210
$endgroup$
$begingroup$
So to prove $xrightarrow{P}$ should I always use $limsup_{nto infty}$ rather than $lim_{nto infty}$
$endgroup$
– SABOY
Jan 10 at 11:59
add a comment |
$begingroup$
We use a $limsup$ because we do not know whether the limit exists, whereas $limsup$ always exists. Note that since your sequence is nonnegative, it goes to zero iff zero is its limsup.
$X$ is a random variable that is as finite. Thus $P(|X| > n)$ is a decreasing sequence with limit $P( forall,n |X| > n)=P(|X|=infty)=0$.
$(*) leq (**)$ is false. However, $(*) leq (***)$ holds.
This is because the event $(|X||Y_n-Y| > epsilon, |X| leq p)$ is contained in the event $(p|Y_n-Y| > epsilon)$, and because the event $(|X||Y_n-Y| > epsilon, |X| > p$) is contained in the event $(|X| > p)$.
Note that for the proof to work, your lower bound on $X$ should be different from the index in the sequence $Y$. The argument reads as follows:
By the same computations, $P(|X||Y_n-Y| > epsilon) leq P(|X| > p)+P(|Y_n-Y| > epsilon/p)$ for every $n,p$ thus the limsup of the LHS is not greater than $P(|X| >p)$ for every $p$. Since this can be arbitrarily close to $0$, the limsup, hence the limit, is $0$.
answered Jan 9 at 22:55
MindlackMindlack
4,770210
$endgroup$
$begingroup$
So to prove $xrightarrow{P}$ should I always use $limsup_{nto infty}$ rather than $lim_{nto infty}$
$endgroup$
– SABOY
Jan 10 at 11:59
add a comment |
$begingroup$
We use a $limsup$ because we do not know whether the limit exists, whereas $limsup$ always exists. Note that since your sequence is nonnegative, it goes to zero iff zero is its limsup.
$X$ is a random variable that is as finite. Thus $P(|X| > n)$ is a decreasing sequence with limit $P( forall,n |X| > n)=P(|X|=infty)=0$.
$(*) leq (**)$ is false. However, $(*) leq (***)$ holds.
This is because the event $(|X||Y_n-Y| > epsilon, |X| leq p)$ is contained in the event $(p|Y_n-Y| > epsilon)$, and because the event $(|X||Y_n-Y| > epsilon, |X| > p$) is contained in the event $(|X| > p)$.
Note that for the proof to work, your lower bound on $X$ should be different from the index in the sequence $Y$. The argument reads as follows:
By the same computations, $P(|X||Y_n-Y| > epsilon) leq P(|X| > p)+P(|Y_n-Y| > epsilon/p)$ for every $n,p$ thus the limsup of the LHS is not greater than $P(|X| >p)$ for every $p$. Since this can be arbitrarily close to $0$, the limsup, hence the limit, is $0$.
answered Jan 9 at 22:55
MindlackMindlack
4,770210
$endgroup$
We use a $limsup$ because we do not know whether the limit exists, whereas $limsup$ always exists. Note that since your sequence is nonnegative, it goes to zero iff zero is its limsup.
$X$ is a random variable that is as finite. Thus $P(|X| > n)$ is a decreasing sequence with limit $P( forall,n |X| > n)=P(|X|=infty)=0$.
$(*) leq (**)$ is false. However, $(*) leq (***)$ holds.
This is because the event $(|X||Y_n-Y| > epsilon, |X| leq p)$ is contained in the event $(p|Y_n-Y| > epsilon)$, and because the event $(|X||Y_n-Y| > epsilon, |X| > p$) is contained in the event $(|X| > p)$.
Note that for the proof to work, your lower bound on $X$ should be different from the index in the sequence $Y$. The argument reads as follows:
By the same computations, $P(|X||Y_n-Y| > epsilon) leq P(|X| > p)+P(|Y_n-Y| > epsilon/p)$ for every $n,p$ thus the limsup of the LHS is not greater than $P(|X| >p)$ for every $p$. Since this can be arbitrarily close to $0$, the limsup, hence the limit, is $0$.
answered Jan 9 at 22:55
MindlackMindlack
4,770210
answered Jan 9 at 22:55
MindlackMindlack
4,770210
answered Jan 9 at 22:55
MindlackMindlack
4,770210
answered Jan 9 at 22:55
MindlackMindlack
4,770210
4,770210
$begingroup$
So to prove $xrightarrow{P}$ should I always use $limsup_{nto infty}$ rather than $lim_{nto infty}$
$endgroup$
– SABOY
Jan 10 at 11:59
add a comment |
$begingroup$
So to prove $xrightarrow{P}$ should I always use $limsup_{nto infty}$ rather than $lim_{nto infty}$
$endgroup$
– SABOY
Jan 10 at 11:59
$begingroup$
So to prove $xrightarrow{P}$ should I always use $limsup_{nto infty}$ rather than $lim_{nto infty}$
$endgroup$
– SABOY
Jan 10 at 11:59
$begingroup$
So to prove $xrightarrow{P}$ should I always use $limsup_{nto infty}$ rather than $lim_{nto infty}$
$endgroup$
– SABOY
Jan 10 at 11:59
add a comment |
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